Given acceleration, a ⃗ = 2 e − t i ^ + 5 c o s t j ^ − 3 s i n t k ^ \vec{a} = 2e^{-t}\hat{i}+5cost\hat{j}-3sint\hat{k} a = 2 e − t i ^ + 5 cos t j ^ − 3 s in t k ^
Velocity is given by, v ⃗ = ∫ a ⃗ d t = ∫ ( 2 e − t i ^ + 5 c o s t j ^ − 3 s i n t k ^ ) d t \vec{v} = \int \vec{a} dt = \int (2e^{-t}\hat{i}+5cost\hat{j}-3sint\hat{k})dt v = ∫ a d t = ∫ ( 2 e − t i ^ + 5 cos t j ^ − 3 s in t k ^ ) d t
Integrating, v ⃗ = ( − 2 e − t + a ) i ^ + ( 5 s i n t + b ) j ^ − ( − 3 c o s t + c ) k ^ \vec{v} = ( -2e^{-t} + a)\hat{i} + (5sint + b)\hat{j} - (-3cost + c)\hat{k} v = ( − 2 e − t + a ) i ^ + ( 5 s in t + b ) j ^ − ( − 3 cos t + c ) k ^
Since we are given velocity at t=0,
Then,
4 i ^ − 3 j ^ + 2 k ^ = ( − 2 e 0 + a ) i ^ + ( 5 s i n 0 + b ) j ^ − ( − 3 c o s 0 + c ) k ^ 4\hat{i}-3\hat{j}+2\hat{k} = ( -2e^{0} + a)\hat{i} + (5sin0 + b)\hat{j} - (-3cos0 + c)\hat{k} 4 i ^ − 3 j ^ + 2 k ^ = ( − 2 e 0 + a ) i ^ + ( 5 s in 0 + b ) j ^ − ( − 3 cos 0 + c ) k ^
Comparing both sides, − 2 + a = 4 ⟹ a = 6 , b = − 3 , c = 1 -2+a = 4 \implies a = 6, b=-3, c=1 − 2 + a = 4 ⟹ a = 6 , b = − 3 , c = 1
Then, v ⃗ = ( − 2 e − t + 6 ) i ^ + ( 5 s i n t − 3 ) j ^ − ( − 3 c o s t + 1 ) k ^ \vec{v} = ( -2e^{-t} + 6)\hat{i} + (5sint -3)\hat{j} - (-3cost + 1)\hat{k} v = ( − 2 e − t + 6 ) i ^ + ( 5 s in t − 3 ) j ^ − ( − 3 cos t + 1 ) k ^
Displacement is given by, s ⃗ = ∫ v ⃗ d t \vec{s} = \int \vec{v} dt s = ∫ v d t
s ⃗ = ∫ [ ( − 2 e − t + 6 ) i ^ + ( 5 s i n t − 3 ) j ^ − ( − 3 c o s t + 1 ) k ^ ] d t \vec{s} = \int [( -2e^{-t} + 6)\hat{i} + (5sint -3)\hat{j} - (-3cost + 1)\hat{k}] dt s = ∫ [( − 2 e − t + 6 ) i ^ + ( 5 s in t − 3 ) j ^ − ( − 3 cos t + 1 ) k ^ ] d t
Integrating, s ⃗ = [ ( 2 e − t + 6 t + d ) i ^ + ( − 5 c o s t − 3 t + f ) j ^ − ( − 3 s i n t + t + g ) k ^ ] \vec{s} = [( 2e^{-t} + 6t+d)\hat{i} + (-5cost -3t+f)\hat{j} - (-3sint + t+g)\hat{k}] s = [( 2 e − t + 6 t + d ) i ^ + ( − 5 cos t − 3 t + f ) j ^ − ( − 3 s in t + t + g ) k ^ ]
d, f and g are constant.
We are given position at time t=0,
1 i ^ − 3 j ^ + 2 k ^ = [ ( 2 e 0 + 6 × 0 + d ) i ^ + ( − 5 c o s 0 − 3 × 0 + f ) j ^ − ( − 3 s i n 0 + 0 + g ) k ^ ] 1\hat{i}-3\hat{j}+2\hat{k} = [( 2e^{0} + 6\times0+d)\hat{i} + (-5cos0 -3\times0 +f)\hat{j} - (-3sin0 + 0+g)\hat{k}] 1 i ^ − 3 j ^ + 2 k ^ = [( 2 e 0 + 6 × 0 + d ) i ^ + ( − 5 cos 0 − 3 × 0 + f ) j ^ − ( − 3 s in 0 + 0 + g ) k ^ ]
Comparing both sides,
we get, d = − 1 , f = 2 , g = − 2 d=-1, f=2, g=-2 d = − 1 , f = 2 , g = − 2
Then Displacement will be,
s ⃗ = ( 2 e − t + 6 t − 1 ) i ^ + ( − 5 c o s t − 3 t + 2 ) j ^ − ( − 3 s i n t + t − 2 ) k ^ \vec{s} = ( 2e^{-t} + 6t-1)\hat{i} + (-5cost -3t+2)\hat{j} - (-3sint + t-2)\hat{k} s = ( 2 e − t + 6 t − 1 ) i ^ + ( − 5 cos t − 3 t + 2 ) j ^ − ( − 3 s in t + t − 2 ) k ^
Comments
This is really so helpful,thankyou